The total payments over 30 years is:
#t = ($450.23)/(month) xx 30 years xx (12 months) / (year)#
#t = ($450.23)/color(blue)(cancel(color(black)(month))) xx 30 color(red)(cancel(color(black)(years))) xx (12 color(blue)(cancel(color(black)(months)))) / color(red)(cancel(color(black)(year)))#
#t = $450.23 xx 30 xx 12#
#t = $450.23 xx 360#
#t = $162082.80#
Now, assuming the $52,000 principle of the loan was paid of during the 30 years we can find the total interests as:
#i = t - p#
Where:
#i# is the total interest paid, what we are solving for in this problem.
#t# is the total payments made - $162,082.80 for this problem.
#p# is the principle paid - $52,000.00 for this problem.
Substituting and solving for #i# gives:
#i = $162,082.80 - $52,000.00#
#i = $110082.80#
